Estimated 3203 Light Bulb

Thirds

L changes in the shadow of two parts. Part I: the wall shadow is 0, the shadow of the growing surface, relatively simple variant, when x takes the maximum value of the first portion (x = l below) when it is taken to a first maximum value L portion; a second section, increasing the shadow of the wall, ground shadow continues to decrease, it changes a little bit more complicated than the first part, but you can still determine a sudden function on a single peak, so it can be resolved by one-third. Here are analyzed for the two parts

Left side wall and is provided to the person's position is x, the first section is a range (0 <= x <= l), is the range of the second portion (l <= x <= D), focus will be subject to conversion to Resolution function l is obtained and a second portion changes:

A first portion (seeking l):

Image taken directly to the maximum value:

Obtained by a simple calculation similar triangles: l = DD * h / H

A second portion (Resolution Function requirements of the second portion);

The shadow of the ground segment was simply Dx, focusing on the shadow on the wall, we are here to set him y

Also, as auxiliary lines, using similar triangles give y = H + D (Hh) / x;

L=D-x+H+D(H-h)/x;

After the two parts are considered good, you can write pointers, A out of this problem!

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int T;
 4 double H,h,D;
 5 double f(double x){
 6     return D-x+H+D*(h-H)/x;
 7 }
 8 int main(){
 9     scanf("%d",&T);
10     for(int X=1;X<=T;X++){
11         scanf("%lf%lf%lf",&H,&h,&D);
12         double l=D-D*h/H,r=D;
13         while(r-l>=1e-3){
14             double m1=l+(r-l)/3,m2=r-(r-l)/3;
15             if(f(m1)<f(m2))l=m1;
16             else r=m2;
17         }
18         printf("%.3lf\n",f(r));
19     }
20     return 0;
21 }